Question

In quadrilater ABCD side BC parallel side AD. Diagonal AC and diagonal BD intersect in point Q. If AQ=1/3AC, then show that DQ=1/2BQ

Answer 1

Answer:

Given: in quadrilateral ABCD,

AD ║ BC  and  AQ=1/3 AC

To prove:  DQ=1/2 BQ

Proof:

Since, AQ=1/3 AC

⇒ AQ/AC = 1/3

Let, AQ = x and AC = 3x

Where x is any number,

⇒ CQ = AC - AQ = 3x - x = 2x

Now, In quadrilateral ABCD,

AD ║ BC

By the Alternative interior angle theorem,

$\angle QAD\cong \angle QCB$,

$\angle QDA\cong \angle QBC$

By AA similarity postulate,

$\triangle ADQ\sim\triangle CBQ$

Now, By the property of similar triangles,

$\frac{DQ}{BQ} = \frac{AQ}{CQ}$

$\frac{DQ}{BQ} = \frac{x}{2x}$

$DQ = \frac{1}{2}BQ$

Hence, Proved.